Answer:
4.76
Explanation:
In this case, we have to start with the <u>buffer system</u>:
We have an acid () and a base (). Therefore we can write the <u>henderson-hasselbach reaction</u>:
If we want to calculate the pH, we have to <u>calculate the pKa</u>:
According to the problem, we have the <u>same concentration</u> for the acid and the base 0.1M. Therefore:
If we divide:
If we do the Log of 1:
So:
With this in mind, the pH is 4.76.
I hope it helps!
Answer:
0.16 J/g℃
Explanation:
We have the following data:
heat = 1,200 J
mass of metal = m = 150 g
change in temperature = ΔT = 50.0℃
The heat absorbed is calculated by using the following equation:
heat = m x Cp x ΔT
So, we introduce the data in the equation to calculate the specific heat of the metal (Cp), as follows:
Cp = heat/(m x ΔT) = (1,200 J)/(150 g x 50.0℃) = 0.16 J/g℃
Answer:
x=-1
Explanation:
1. Simplify brackets.
2x+3×6x=−20
2. Simplify 3×6x to 18x.
2x+18x=−20
3. Simplify 2x+18x to 20x.
20x=−20
4. Divide both sides by 20.
x=−1